As an exercise, I was given the following matrix: $$A=\left[ \begin{array}{cc} 1 & i & \\ 3i & 2 & i & \\ & 3i & 3 & i & \\ & & \ddots & \ddots & \ddots & \\ & & & 3i & n-1 & i \\ & & & & 3i & n \end{array} \right]$$
I need to locate the eigenvalues of $Re(A)$ using Gershgorin's theorems.
First of all, i calculated $Re(A)$ using the formula $Re(A)=\frac{A+A^*}{2}$, which gives me the following matrix:
$$ \operatorname{Re}(A)= \left[ \begin{array}{cccccc} 1 & -i & & & & \\ i & 2 & -i & & & \\ & i & 3 & -i & & \\ & & \ddots & \ddots & \ddots & \\ & & & i & n-1 & -i\\ & & & & i & n \end{array} \right] $$
And got the following disks: $$ H_1=(1,1)\\H_2=(2,2)\\H_3=(3, 2)\\.\\.\\.\\H_{n-1}=(n-1, 2)\\H_n=(n,1)\\ $$
Using the first Gershgorin's theorems I can say that the each eigenvalue $i$ is in the range $0\le \lambda_i \le n+1$.
This is an Hermitian matrix, so its eigenvalues must be real. It's also a strictly diagonally dominant matrix, which means that it can't be invertible, so $0$ can't be an eigenvalue of $Re(A)$, which menas that each eigenvalue $i$ is in the range $0< \lambda_i \le n+1$.
As a bonus point, since this matrix is hermitian and it's eigenvalues are all real and positive (because $0$ can't be an eigenvalue) it also means that the matrix $A$ is positive definite because $Re(A)$ is positive definite (when studying the Sylvester's criterion in my course, I read that if a matrix is not hermitian, I can apply Sylvester's criterion to $Re(A)$ instead to prove that $A$ is positive definite.
I'm not sure my assumptions are right, I just want to make sure that what i said isn't wrong. Did I make mistakes here? Thank you in advance!
